Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 15 - Think and Solve: 38


a. Find the amount of heat needed for steel. $$Q = cm \Delta T$$ $$=\frac{450 J}{kg \cdot ^{\circ}C}(10 kg)(100^{\circ}C - 0^{\circ}C ) = 450,000 J$$ b. Now perform the calculation for water. $$Q = cm \Delta T$$ $$=\frac{4190 J}{kg \cdot ^{\circ}C}(10 kg)(100^{\circ}C - 0^{\circ}C ) = 4,190,000 J$$

Work Step by Step

Heating the water through a certain temperature difference takes almost ten times the energy that it does for steel. This is not surprising, because water has a large specific heat capacity.
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